Problem: $\overline{AC}$ is $3$ units long $\overline{BC}$ is $4$ units long $\overline{AB}$ is $5$ units long What is $\csc(\angle ABC)?$ $A$ $C$ $B$ $3$ $4$ $5$
Solution: $\csc(\angle ABC) = \dfrac{1}{\sin(\angle ABC)}$ How can we find $\sin(\angle ABC)$ SOH CAH TOA in = pposite over ypotenuse Opposite $= \overline{AC} = 3$ Hypotenuse $= \overline{AB} = 5$ $\sin(\angle ABC) = \dfrac{3}{5}$ $\csc(\angle ABC) = \dfrac{1}{\sin(\angle ABC)} = \dfrac{5}{3}$